Simplifying an Equation with Taylor Series Expansion

[hermano]

Homework Statement

In a paper I have found there is a simplification of an equation which I do not understand.
How did they do this? Can someone explain this step by step?

Homework Equations

The euqation is as follows:

X = e \cos\theta + \sqrt{(R^2 + e^2 \sin^2 \theta)}

X = distance
e = eccentricity
theta = angle
R = radius

Since the eccentricity is much than 1000 times smaller than the radius the above equation is simplified to:

X = e \cos\theta + R + \frac{e^2}{2R} \sin^2 \theta

The Attempt at a Solution

In the paper they say that they used the Taylor series expansion in terms of \sin^2 \theta for this simplification.

 

[DrClaude]

The Taylor expansion being made is that of $(1+x)^{1/2}$ with $|x| \ll 1$:
$$
\begin{align}
(R^2 + e^2 \sin^2 \theta)^{1/2} &= R \left( 1 + \frac{e^2}{R^2} \sin^2 \theta \right)^{1/2} \\
&\approx R \left( 1 + \frac{e^2}{2 R^2} \sin^2 \theta \right) \\
&= R + \frac{e^2}{2 R} \sin^2 \theta
\end{align}
$$

 

[hermano]

Thank you DrClaude. Can you also help me to simplify the following equation using the Taylor series:

(R^2-e^2 \sin^2 \theta)^{1/2}

 

[NascentOxygen]

Thank you DrClaude. Can you also help me to simplify the following equation using the Taylor series:

(R^2-e^2 \sin^2 \theta)^{1/2}

Follow the example DrClaude has given, except now replace every x by (-x) in the Taylor expansion:

(1+x)^n = 1 + (n)(x) + (n)(n-1)(x^2)/2! + ...

 

[hermano]

Follow the example DrClaude has given, except now replace every x by (-x) in the Taylor expansion:

(1+x)^n = 1 + (n)(x) + (n)(n-1)(x^2)/2! + ...

f(x) = (1-x)^(1/2) -> f(0)= 1
f'(x) = (1/2)(1-x)^(-1/2) -> f'(0) = 1/2

So the Tyalor expansion will become:
(1-x)^(1/2) = 1 + (1/2) (x)

and not
(1-x)^(1/2) = 1 - (1/2) (x) as when I replace x by (-x) in the Taylor expansion for (1+x)^(1/2)

Or what do I wrong?

 

[DrClaude]

f(x) = (1-x)^(1/2) -> f(0)= 1
f'(x) = (1/2)(1-x)^(-1/2) -> f'(0) = 1/2

There is a minus sign missing:
$$
\begin{align}
\frac{d}{dx} (1-x)^{1/2} &= \frac{1}{2} (1-x)^{-1/2} \frac{d}{dx} (1-x) \\
&= -\frac{1}{2} (1-x)^{-1/2}
\end{align}
$$

 

[hermano]

Thanks! That's the error I have made.